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Question: Hall Heroult process is given by: A.\[C{{r}_{2}}{{O}_{3}}+2Al\to A{{l}_{2}}{{O}_{3}}+2Cr\] B.\[C...

Hall Heroult process is given by:
A.Cr2O3+2AlAl2O3+2CrC{{r}_{2}}{{O}_{3}}+2Al\to A{{l}_{2}}{{O}_{3}}+2Cr
B.Cu2+(aq)+H2(g)Cu(s)+2H+(aq)C{{u}^{2+}}(aq)+{{H}_{2}}(g)\to Cu(s)+2{{H}^{+}}(aq)
C.ZnO+CZn+CO(Coke, 1673 K)ZnO+C\to Zn+CO\,(\text{Coke, 1673 K})
D.2Al2O3+3C4Al+3CO22A{{l}_{2}}{{O}_{3}}+3C\to 4Al+3C{{O}_{2}}

Explanation

Solution

To answer this question, you should recall the concept of Hall-Heroult’s process. This is used in the extraction of aluminium from its ore.

Complete step by step answer:
The electrolysis of alumina through the process of Hall and Heroult is carried by utilising a fusion mixture of alumina and cryolite in the presence of minor quantities of aluminium fluoride and fluorspar. The addition of cryolite and fluorspar leads to an increase in the electrical conductivity of alumina and lowers the fusion temperature.
At the anode, alumina reacts with fluorine to give oxygen. The oxygen gas released undergoes a chemical reaction with carbon to form COCO and CO2C{{O}_{2}}​. These gases are liberated at the anode. Pure Al2O3A{{l}_{2}}{{O}_{3}} is mixed with CaF2 or Na3AlF6Ca{{F}_{2}}~or\text{ }N{{a}_{3}}Al{{F}_{6}}which is known as cryolite. The role of cryolite is an aid in lowering the melting point of the mixture and increases its ability to conduct electricity.
The overall reaction is:
2Al2O3 + 3C  4Al + 3CO2\mathbf{2A}{{\mathbf{l}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{3}}}~+\text{ }\mathbf{3C}\text{ }\to ~\mathbf{4Al}\text{ }+\text{ }\mathbf{3C}{{\mathbf{O}}_{\mathbf{2}}}.

Hence, the correct option is D.

Note:
Aluminium is a highly reactive metal, belonging to the IIIA group of the periodic table. In nature, aluminium is generally found as oxides. Some of these few important oxide ores include:
Bauxite Al2O3.2H2OA{{l}_{2}}{{O}_{3}}.2{{H}_{2}}O
Corundum Al2O3A{{l}_{2}}{{O}_{3}}
Cryolite Na3AlF6N{{a}_{3}}Al{{F}_{6}}
Aluminium is mostly extracted from its bauxite ore. The bauxite ore contains ferric oxide and silica as impurities. It is first concentrated by gravity separation of ferric oxide impurities. As these impurities are magnetic, they are separated by the process of magnetic separation. The ore is then concentrated by the chemical process. Aluminium ions are created at the adverse cathode from the aluminium oxide and then sink because they are heavier than the cryolite solution. Then, the liquid shape of the aluminium that has sunk to the bottom. On the other side, at the positive anode, the oxygen from the aluminium oxide forms and responds to carbon dioxide CO2C{{O}_{2}}with the graphite carbon.